Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here: math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world — including you! You’ll find this week’s puzzle below.
Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the link below. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday — have a great weekend!
Before we get to the new puzzle, let’s return to last week’s. Congratulations to 👏 James Ledoux 👏 of Litchfield, New Hampshire, our big winner. You can find a solution to the previous Riddler at the bottom of this post.
Now here’s this week’s Riddler, a tenure-track puzzle that comes to us from Larry Denenberg from Newton, Massachusetts.
A university has 10 mathematicians, each one so proud that, if she learns that she made a mistake in a paper, no matter how long ago the mistake was made, she resigns the next Friday. To avoid resignations, when one of them detects a mistake in the work of another, she tells everyone else but doesn’t inform the mistake-maker. All of them have made mistakes, so each one thinks only she is perfect. One Wednesday, a super-mathematician, whom all respect and believe, comes to visit. She looks at all the papers and says: “Someone here has made a mistake.”
What happens then? Why?
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Need a hint? You can try asking me nicely. Want to submit a new puzzle or problem? Email me.
And here’s the answer to last week’s Riddler, concerning some mutinous, game-theoretic pirates who are trying to divide up a booty of gold pieces, which came to us from Kyle Joecken.
Since no captain wants to die and everyone is looking out for himself, the game will end after the pirates vote to accept the first proposal, wherein the pirates, from the captain on down to the lowest-ranking pirate, receive 6, 0, 1, 0, 1, 0, 1, 0, 1 and 0 pieces of gold, respectively. No blood is shed.
Why? Let’s work backward. Give each crew member a name: P1 (the original captain) down to P10 (the lowest-ranking pirate). Imagine we’ve reached the situation where there are only two pirates, P9 and P10, left alive. (This won’t happen in practice, but the outcome of this counterfactual scenario will inform pirates’ decisions.) Since it only takes half the votes to approve a proposal, P9 can propose whatever he likes and it’ll be approved. So P9, in this case the acting captain, proposes he gets all 10 gold pieces, and P10 gets nothing. Now imagine, instead, there are three living pirates. P8 is the captain now, and he doesn’t want to die, so he needs to win the vote of at least one of the two other pirates. The cheapest way to do this is to offer one piece of gold to P10 and nothing to P9, leaving nine for himself. P10 will vote in favor of this proposal because, as we just saw, he’ll get no gold if P8 dies.
Now imagine there are four living pirates. P7 is the captain now and needs to persuade one other pirate to vote for his plan. The cheapest way to do this is to give one piece of gold to P9, who gets nothing if only three pirates remain, so he will vote yes, and then P7 gets to keep the rest for himself. Let’s do just one more: Suppose there are five living pirates. P6 is the captain in this case, and he needs to persuade two other pirates to vote for his plan, lest he walk the plank. If P6 dies and only four pirates remain, P8 and P10 will get no gold, as we just saw. So P6 can bribe P8 and P10 with one piece each, keeping the rest for himself, for a ranked division of 8-0-1-0-1. (Remember, these pirates are well-educated in game theory. They’re very good at describing and understanding optimized scenarios.)
Extending this logic back to the original case of 10 pirates, the game ends with an initially proposed division of 6-0-1-0-1-0-1-0-1-0.
What about the extra-credit, generalized problem with P pirates and G pieces of gold? If there’s enough gold (specifically, if \(G \geq P/2 – 1\)) the pattern we just found will still hold. The captain will keep all the gold minus the one-gold-piece bribes alternating down the chain of command. (This includes the case where the captain keeps zero gold pieces, but also keeps his life.) If there is not enough gold (if \(G<P/2 -1\)), the captain can’t afford the bribes he needs, and there will be blood.
Things get a little trickier (and a lot messier) in this latter case. From Riddler Hall of Famer Laurent Lessard: If \(P>2G+2\), then the captains will die until there are \(P=2G+2^k\) pirates remaining with k as large as possible. Then, pirates 2, 4, 6, … , 2G each get one gold piece, and nobody else gets anything. However, in this case the solution is not unique, in the sense that once we arrive at \(P=2G+2^k\), there are other admissible ways of dividing up the gold besides the one described above.
Also see these approaches from Geoff Buchan, Sujeet Akula and Tim Urban from the blog Wait But Why.
Andrew Mascioli sums it up well: “The moral of the story seems to be that if you have enough money to buy some friends, being the captain is a good deal! But the alternative brings out the problems with having a bloodthirsty crew.”
One reader proposed adopting a handsome new pirate flag based on this puzzle’s solution:
Elsewhere in the puzzling world:
- A fantasy author hid some treasure to promote his book. “But 35 years later, people are still searching.” [Vice]
- Buried in the card game Set is the treasure of some deep combinatorial mathematics. [Quanta Magazine]
- A jelly bean jar puzzle with a delicious solution. [The New York Times]
- The puzzle of the sliding blocks. [The Guardian]
- As the hockey season winds down, some hockey problems! [Expii]
Have a lovely weekend!